Re: Basic Calculus Question
- From: Kira Yamato <no@xxxxxxxx>
- Date: Wed, 22 Jun 2005 03:35:00 GMT
Robert Israel wrote:
In article <DGPte.8905$pa3.7913@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>, Kira Yamato <no@xxxxxxxx> wrote:
I'm having problem with the following question:
Suppose (1) f in C^2(R), (2) |f(x)| < 1 for all x in R, (3) [f(0)]^2 + [f'(0)]^2 = 2^2. Show that there exists p in R such that f''(p) + f(p) = 0.
My attempt: There is a hint to investigate g(x) := [f(x)]^2 + [f'(x)]^2,
from which I deduce
(4) g'(x) = 2f'(x)[f''(x)+f(x)].
Because of (3), f' cannot be identically zero. So, by continuity of f', there exists a neigborhood around x=0 such that
f'(x) != 0.
But I do not know how to show that within this neigborhood that there must exists a point p such that
g'(x) = 0.
This is somewhat more subtle.
Outline:
Note that in any interval where g(x) >= 3, f'(x) can't be too close to 0
(and its sign must be constant). Therefore such an interval can't
be too large. In the largest such interval containing 0, g must
have a maximum.
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
Ah... Yes. In particular, this interval must be bounded or otherwise f would fail to be bounded. Then, by Rolle's Theorem, there exists p such that g'(p) = 0.
Thanks. -kira .
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- From: Kira Yamato
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